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Oscillation question

  1. Mar 20, 2015 #1
    1. The problem statement, all variables and given/known data

    262tgef.jpg

    2. Relevant equations

    x = x₀cos(wt) (or x = -x₀cos(wt) , depending on the graph shape)
    x = x₀sin(wt)
    ω = 2π / T


    3. The attempt at a solution

    vxf606.jpg

    I am confused as I have never encountered any oscillatory object with a thickness. Because if I take the middle of the pathway to be the equilibrium point, the middle of the blade would not touch the ceiling/floor of the pathway, so it can't be considered as a full oscillation. Which was why I assumed the oscillating object was a 'line', and split the 0.4m blade into 2 'blocks' of 0.4m on the top and bottom of the pathway

    What did I do wrong?
     
  2. jcsd
  3. Mar 20, 2015 #2

    Suraj M

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    The ##t₁ =0.4259## that you have found is the time for the blade to come down to 0.55m below equilibrium from eq.,
    you have the right idea, you have tried to find the time the blade takes to get up from the base to 0.55 m (below equilibrium) but the t₁ you calculated is not what you want, try finding the real t₁ value you need from the total time and the one you have calculated.
     
  4. Mar 20, 2015 #3
    Oh, so just to clarify, for the ##x = -x₀cos(wt)## formula, if I substitute a value of ##x##, the ##t## that I find would refer to the time taken for the object to travel from the equilibrium point to the displaced point?

    But even if the t1 refers to the time taken to travel from the equilibrium point to -0.55 displacement, the final answer would then be ##4/2## (half the period) + ##2*0.4259## (refer to the yellow portion of the graph I have drawn) = ##2.85##, which is still not the correct answer (2.2).
     
  5. Mar 20, 2015 #4

    Suraj M

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    i correct myself, everywhere i said .55 m below equilibrium, it should be .15 m below equilibrium.
    and actually forget whatever i said earlier.
    As you have used cos function the time you actually get is the time it takes for the body to rise from bottom to 0.55 m off the ground. as you had stated earlier. :) i'm sorry, i thought you were writing ##\sin## everywhere.
    but you have to use ##x = 0.15 m## (as the point 0.55 m off the ground is 0.15 m below equilibrium point )and not 0.55 m
    by this method, you'll get 2.2 s
     
    Last edited: Mar 20, 2015
  6. Mar 20, 2015 #5
    Oh okay. But if you refer to the diagram I drew on the left, I assumed that there's a 0.4m block of blade on the ground, such that the oscillation has an amplitude of only 0.7m. So 0.55m off the ground would mean 0.15m above the 'block'. So that means 0.15m above the minimum point of oscillation (ie. displacement = -0.55m).

    Weird... or is my diagram/understanding of how the blade oscillates wrong in the first place?
     
  7. Mar 20, 2015 #6

    haruspex

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    I can't follow your calculation there. Mr. Jones only cares about the bottom edge of the blade. Treat that as the oscillating entity.
     
  8. Mar 21, 2015 #7
    That makes sense - I got the correct answer. Thanks a lot guys! :woot:
     
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